8 Answers
8

Here's a semi-explicit construction for a smooth function f that is zero precisely on the classical Cantor set. By this set I mean the one that is obtained from I_0 = [0,1] by repeatedly removing the middle third of any ensuing interval. So let's denote by I_n the
n-th set in this process.

Now let's make a smooth function f_n on [0,1] such that its zero set is exactly I_n.
Starting with f_0 = 0 we obtain f_{n+1} from f_n as follows:

set f_{n+1} = f_n on I_{n+1} and
on an interval that is removed from I_n make f_{n+1} equal to a bump function
that is 0 only at the boundary of the interval. We can choose the bump function to be of height 2^{-2^n}.

This choice of heights of the bump functions will ensure that
the derivatives of f all converge uniformly to their pointwise limits.
Hence the limit function f_n is again smooth. By construction its zero set is exactly
the Cantor set.

If you work through the details, I suspect you may find you need the bumps at step $n$ to have heights decreasing faster than $2^{-n}$. The reason is that the bump has to fit into an interval of length $3^{-n}$, and the resulting squeeze makes the derivatives big (relatively speaking).
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Harald Hanche-OlsenMay 9 '10 at 19:03

Each bump in the bump function has width 1/3^n. If the height is h(n), then the largest value of the kth derivative of of the stage n bumps will be 3^{kn}h(n) times the largest value of the kth derivative of the original bump function. If you want these to converge uniformly, the exact condition you need is that 3^{kn}h(n) converges to 0. So the necessary and sufficient condition on h to make the construction work is that for all a>0, h(n) is eventually less than a^n. So yes, 2^{-2^n} will work.
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David DiamondstoneMay 16 '11 at 18:24

I admit I have never worked through the proof of this result. So I wonder, why partitions of unity? It seems to me that the local result is no easier than the global result.
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Harald Hanche-OlsenMay 9 '10 at 19:06

One can avoid using the partition of unity for that, just summation of "1-bump" functions with suitable coefficients..
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PetyaMay 10 '10 at 3:22

Petya, I'm sure you're correct, but it's a bit more fiddly to ensure the uniform convergence of the sum of the derivatives of the bump functions without a convenient global coordinate system.
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Robin ChapmanMay 10 '10 at 9:07

I can't resist trying my hand at sketching a proof of the general result given in Robin Chapman's answer. Let $F\subset\mathbb{R}^n$ be any closed set. Let $E_0=\{x\colon\operatorname{dist}(x,F)\ge1\}$, and for $k=1,2,\ldots$ let $E_k=\{x\colon2^{-k}\le\operatorname{dist}(x,F)\le2^{1-k}\}$. Let $\omega$ be a standard mollifier, and put $$f=\sum_{k=0}^\infty \alpha_k\chi_{E_k}*\omega_k,\qquad\omega_k(y)=2^{nk}\omega(2^ky),$$ where $\alpha_k>0$ decays fast enough so all derivatives converge uniformly ($\alpha_k=2^{-k^2}$ ought to be sufficient).

Here's an answer from probability: a Brownian motion $B_t$ is a random, continuous function whose zero set is closed, nowhere dense, and has no isolated points. That is, $\{t : B_t = 0 \}$ is almost-surely a topological Cantor set (see, for example, Section 8 of Lalley's lecture notes).

To mention a further point not covered in existing answers: while any closed subset of $\mathbb{R}$ can be the zero set of a smooth function, the zero set of an analytic function either consists entirely of isolated points, or is all of $\mathbb{R}$. To see this we note that if the zero set of an analytic function $f$ contains an accumulation point, then by taking a power series expansion of $f$ at the accumulation point we may extend $f$ locally to a small complex disc around that point, and apply the Identity Theorem from complex analysis to show that $f$ is everywhere zero within that disc. In particular the zero set contains an open neighbourhood in $\mathbb{R}$ of the accumulation point, and using connectedness we can repeat this argument to show that the zero set must be all of $\mathbb{R}$.

André Henriques answer for a closed set $C$ can easily be improved to $C^\infty$ by considering $e^{1/(\alpha-x)+1/(x-\beta)}$ if $x\not\in C$ where
$\alpha$ is the supremum of all elements $< x$ in $C$ (and $\alpha=-\infty$ if $C$ contains no elements which are $< x$) and where similarly $\beta$ is the infimum of all elements $> x$ in $C$ (respectively $\beta=\infty$ if $C$ contains no elements $> x$).